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## Re: [Axiom-math] newbie type problem

**From**: |
Martin Rubey |

**Subject**: |
Re: [Axiom-math] newbie type problem |

**Date**: |
Mon, 20 Dec 2004 17:59:58 +0100 |

Dear Frank,
>* We shall implement derivation rules via rules. First intention is:*
>* ablrules := rule abl(a+b,c) == abl(a,c) + abl(b,c)*
>* *
>* but axiom evaluates abl(a+b,c) to zero and makes a rule 0 ->*
>* abl()+abl(). Simple ...rule 'abl(... does not work because of wrong*
>* type... *
You'd have to read the section "Rules and Pattern Matching" in the axiom
book. The answer is:
abl := operator 'abl
ablplus := rule abl(a+b,c) == abl(a,c) + abl(b,c)
There is also a possibility of "adorning" pattern variables -- see the same
section. In your case, you want to have a rule that applies only to monomials,
so it will be roughly
ablmonomial := rule abl(m | monomial?(m)) == ...
However, it is pretty obvious that rules should only be the quick and dirty
approach...
Hope this helps,
Martin